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Chapter 18: Problem 24

Many elements have been synthesized by bombarding relatively heavy atoms with high-energy particles in particle accelerators. Complete the following nuclear equations, which have been used to synthesize elements. a. \(\quad+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow$$\quad$$+6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+$$\quad$$\rightarrow \frac{260}{105} D b+4 \frac{1}{6} n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \frac{257}{153} \mathrm{Lr}+\)__________

Short Answer

Expert verified
The completed nuclear reactions are: a. \(^{239}_{95}Am+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow\ ^{244}_{98}Cf +6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+^{11}_{7}N \rightarrow \ ^{260}_{105} D b+4 \frac{1}{6}n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \ ^{257}_{153} \mathrm{Lr}+2_{0}^{1}n\)

Step by step solution

01

Find the atomic and mass numbers of the missing element

We need to find the atomic and mass numbers for the missing element in the reaction. Since we know the atomic and mass numbers of other particles involved in the reaction, we can use conservation of atomic and mass numbers to find the missing element. To conserve the atomic numbers, we must have: missing element's atomic number + 2 = 97 To conserve the mass numbers, we must have: missing element's mass number + 4 = 243 Now, let's solve these equations to find the missing element's atomic and mass numbers.
02

Solve the equations

Solving the equations from Step 1, we get the missing element's atomic and mass numbers as: missing element's atomic number = 97 - 2 = 95 missing element's mass number = 243 - 4 = 239 Thus, the missing element in the reaction is \(^{239}_{95}Am\). The completed reaction for (a) is: \(a. \quad ^{239}_{95}Am+\frac{4}{2} H e \rightarrow 243 B k+\frac{1}{0} n\) b. \(^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow\quad+6_{0}^{1} n\)
03

Find the atomic and mass numbers of the missing element

Following the same procedure as before, we need to conserve the atomic and mass numbers in the reaction. To conserve the atomic numbers, we must have: 92 + 6 = missing element's atomic number To conserve the mass numbers, we must have: 238 + 12 = missing element's mass number + 6
04

Solve the equations

Solving the equations from Step 3, we get the missing element's atomic and mass numbers as: missing element's atomic number = 92 + 6 = 98 missing element's mass number = 238 + 12 - 6 = 244 Thus, the missing element in the reaction is \(^{244}_{98}Cf\). The completed reaction for (b) is: \(b. \quad ^{238} \mathrm{U}+^{12}_{6} \mathrm{C} \rightarrow\ ^{244}_{98}Cf +6_{0}^{1} n\) c. \(^{249} \mathrm{Cf}+\quad\rightarrow \frac{260}{105} D b+4 \frac{1}{6}n\)
05

Find the atomic and mass numbers of the missing element

Following the same procedure, we need to conserve the atomic and mass numbers in the reaction. To conserve the atomic numbers, we must have: 98 + missing element's atomic number = 105 To conserve the mass numbers, we must have: 249 + missing element's mass number = 260
06

Solve the equations

Solving the equations from Step 5, we get the missing element's atomic and mass numbers as: missing element's atomic number = 105 - 98 = 7 missing element's mass number = 260 - 249 = 11 Thus, the missing element in the reaction is \(^{11}_{7}N\). The completed reaction for (c) is: \(c. \quad ^{249} \mathrm{Cf}+^{11}_{7}N \rightarrow \ ^{260}_{105} D b+4 \frac{1}{6}n\) d. \(^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \frac{257}{153} \mathrm{Lr}+\text{__________}\)
07

Find the number of neutrons emitted

Since we are given both atomic and mass numbers for both elements in the reaction, we can directly find the number of neutrons emitted by conserving the mass numbers. To conserve the mass numbers, we must have: 249 + 10 = 257 + missing neutrons
08

Solve for the number of neutrons emitted

Solving the equation from Step 7, we get the missing neutrons as: missing neutrons = 249 + 10 - 257 = 2 The completed reaction for (d) is: \(d. \quad ^{249} \mathrm{Cf}+^{10}_{5} \mathrm{B} \rightarrow \ ^{257}_{153} \mathrm{Lr}+2_{0}^{1}n\)

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Most popular questions from this chapter

One type of commercial smoke detector contains a minute amount of radioactive americium-241 ( \(\left.^{(241} \mathrm{Am}\right),\) which decays by \(\alpha\) -particle production. The \(\alpha\) particles ionize molecules in the air, allowing it to conduct an electric current. When smoke particles enter, the conductivity of the air is changed and the alarm buzzes. a. Write the equation for the decay of \(^{241}_{95} \mathrm{Am}\) by \(\alpha\) -particle production. b. The complete decay of \(^{241} \mathrm{Am}\) involves successively \(\alpha, \alpha\) \(\boldsymbol{\beta}, \alpha, \alpha, \boldsymbol{\beta}, \alpha, \alpha, \alpha, \boldsymbol{\beta}, \alpha,\) and \(\boldsymbol{\beta}\) production. What is the final stable nucleus produced in this decay series? c. Identify the 11 intermediate nuclides.

The rate constant for a certain radioactive nuclide is \(1.0 \times\) \(10^{-3} \mathrm{h}^{-1} .\) What is the half-life of this nuclide?

The most significant source of natural radiation is radon-222. \(^{222} \mathrm{Rn},\) a decay product of \(^{238} \mathrm{U},\) is continuously generated in the earth's crust, allowing gaseous Rn to seep into the basements of buildings. Because \(^{222} \mathrm{Rn}\) is an \(\alpha\) -particle producer with a relatively short half-life of 3.82 days, it can cause biological damage when inhaled. a. How many \(\alpha\) particles and \(\beta\) particles are produced when \(^{238} \mathrm{U}\) decays to \(^{222} \mathrm{Rn} ?\) What nuclei are produced when \(^{222} \mathrm{Rn}\) decays? b. Radon is a noble gas so one would expect it to pass through the body quickly. Why is there a concern over inhaling \(^{222} \mathrm{Rn} ?\) c. Another problem associated with \(^{222} \mathrm{Rn}\) is that the decay of \(^{222} \mathrm{Rn}\) produces a more potent \(\alpha\) -particle producer \(\left(t_{1 / 2}=\right.\) 3.11 min) that is a solid. What is the identity of the solid? Give the balanced equation of this species decaying by \(\alpha\) particle production. Why is the solid a more potent \(\alpha\) -particle producer? d. The U.S. Environmental Protection Agency (EPA) recommends that \(^{222}\) Rn levels not exceed 4 pCi per liter of air (1 \(\mathrm{Ci}=1\) curie \(=3.7 \times 10^{10}\) decay events per second; \(1 \mathrm{pCi}=1 \times 10^{-12} \mathrm{Ci}\). Convert \(4.0 \mathrm{pCi}\) per liter of air into concentrations units of \(^{222} \mathrm{Rn}\) atoms per liter of air and moles of \(^{222}\) Rn per liter of air.

Write balanced equations for each of the processes described below. a. Chromium- \(51,\) which targets the spleen and is used as a tracer in studies of red blood cells, decays by electron capture. b. Iodine-131, used to treat hyperactive thyroid glands, decays by producing a \(\beta\) particle. c. Phosphorus- \(32,\) which accumulates in the liver, decays by \(\beta\) -particle production.

A chemist studied the reaction mechanism for the reaction $$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)$$ by reacting \(\mathrm{N}^{16} \mathrm{O}\) with \(^{18} \mathrm{O}_{2}\). If the reaction mechanism is $$\begin{aligned} \mathrm{NO}+\mathrm{O}_{2} & \rightleftharpoons \mathrm{NO}_{3}(\text { fast equilibrium }) \\ \mathrm{NO}_{3}+\mathrm{NO} & \longrightarrow 2 \mathrm{NO}_{2}(\text { slow }) \end{aligned}$$ what distribution of \(^{18} \mathrm{O}\) would you expect in the \(\mathrm{NO}_{2} ? \)Assume that \(\mathrm{N}\) is the central atom in \(\mathrm{NO}_{3},\) assume only \(\mathrm{N}^{16} \mathrm{O}^{18} \mathrm{O}_{2}\) forms, and assume stoichiometric amounts of reactants are combined.
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